We can use the conservation of momentum and the conservation of kinetic energy to solve this problem.
First, let's find the initial momentum of the system:
pinitial = m1*v1 + m2*v2
pinitial = (1.6 kg)(4 m/s) + (2.1 kg)(-2.5 m/s)
pinitial = 1.6 kg*m/s - 5.25 kg*m/s
pinitial = -3.65 kgm/s
Since the collision is elastic, the total kinetic energy of the system is conserved:
KE_initial = (1/2)m1v1^2 + (1/2)m2v2^2
KE_initial = (1/2)(1.6 kg)(4 m/s)^2 + (1/2)(2.1 kg)(-2.5 m/s)^2
KE_initial = 12.8 J + 6.56 J
KE_initial = 19.36 J
Now, let's find the final velocities of the two blocks:
p_final = m1v1' + m2v2'
KE_final = (1/2)m1(v1')^2 + (1/2)m2(v2')^2
Using the conservation of momentum, we can write:
p_final = p_initial
m1v1' + m2v2' = -3.65 kgm/s
Solving for v2', we get:
v2' = (-m1v1' - 3.65 kgm/s) / m2
Using the conservation of kinetic energy, we can write:
KEfinal = KEinitial
(1/2)m1(v1')^2 + (1/2)m2(v2')^2 = 19.36 J
Substituting v2' from the momentum equation, we get:
(1/2)m1(v1')^2 + (1/2)m2((-m1v1' - 3.65 kgm/s) / m2)^2 = 19.36 J
Simplifying and solving for v1', we get:
v1' = 1.5 m/s
Substituting v1' into the momentum equation, we get:
v2' = -0.85 m/s
Therefore, the final velocities of the two blocks after the collision are v1' = 1.5 m/s to the right and v2' = -0.85 m/s to the left.
May 12, 2023